Feelin' sixey

All we have to do is just to find the "average" of this function and we can do it without using "really" integrate

first we change this function to other form

$\sin^{6}(\theta)=(\sin^{2}(\theta))^{3}$

$(\frac{1-\cos(2\theta)}{2})^{3}=\frac{1-3\cos(2\theta)+3\cos^{2}(2\theta)-\cos^{3}(2\theta)}{8}$

$\frac{1}{8}(1-3\cos(2\theta)+\frac{3}{2}(1+\cos(4\theta))+\frac{-3\cos(2\theta)-\cos(6\theta)}{4})$

and we know that average of sine and cosine function is 0,so

$<sin^{6}(\theta)>=\frac{1}{8}(\frac{5}{2})=\frac{5}{16}$

then $5+16=21 ..Ans$

We're trying to find

$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=0}^{n} \sin^{6}(\frac{i\pi}{n})$

This is a Riemann sum, so we express it in integral form:

$\frac{1}{\pi} \int_{0}^{\pi} \sin^{6} x \ dx = \frac{1}{\pi} \frac{1 \times 3 \times 5 \times \pi}{2 \times 4 \times 6} = \boxed{\frac{5}{16}}$

The catch is integrating (sinx)^6 from 0 to pi......this can be done easily using Beta Function.......